Do the transpose operator and the adjoint operator have same ranges and kernels?

Question

Let $A$ be a bounded operator on a comlex Hilbert space $H$.
Do we have :

$R(A^t)=R(A^*)$

And

$Ker(A^t)=Ker(A^*)$ ?

Thank you!

Answer 1

Let $A=\begin{bmatrix} 1&i\\0&0\end{bmatrix}$. Then $$ R(A^*)=\left\{\begin{bmatrix} x\\ -ix\end{bmatrix}:\ x\in\mathbb C\right\}, $$ while $$ R(A^t)=\left\{\begin{bmatrix} x\\ ix\end{bmatrix}:\ x\in\mathbb C\right\}. $$ You can see that they are not the same, since $\begin{bmatrix}1\\ i\end{bmatrix}$ is in $R(A^t)$ but not in $R(A^*)$.

For the kernels, both $A^t$ and $A^*$ give a counterexample.